好问题:)
我想出了一种在 O(n * 2^n * 31 * n) 中运行的方法，对于 n = 15，对于一个测试用例来说它有点慢 (228556800)。以下是详细信息：
我这里使用的是dp方法(memoization)，我们定义一个状态为(int mask, int pos)：
我们从最高位到最低位开始，每次检查给定数字的当前位设置的计数，我们将其表示为one_cnt，如果
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one_cnt 是偶数
当前 pos 的异或为零，我们只是移动到 (mask, pos - 1)
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one_cnt 是奇数
如果 one_cnt 等于 n（全奇数），这里我们认为是一个坏状态，什么也不做。否则我们迭代在 pos 处包含零的数字并尝试在此处放置一个。
请注意，当 one_cnt 完全为奇数时，我们将其视为不良状态，因为我们不想增加到可能影响先前状态的 (pos + 1)（违反 dp 原则）。
但是会有这样的情况：arr = [1, 1, 1] 并且解决方案存在。所以在这里我们尝试做一些额外的计算：
我们从最高位 pos 开始，检查当前位是否包含甚至一位，如果是，我们迭代数字以将 1 设置为当前 pos 中为零的数字，然后我们开始我们的记忆并更新我们的结果。
例如如果 arr = [1, 1, 1]，我们可以检查 [2, 1, 1], [1,2,1], [1,1,2]
希望我已经解释清楚了。
如果我想出更快的方法，我会更新解决方案:)
代码如下：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <ctime>
#include <cassert>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define range(i, n) for (long long i=0; i<(n); ++i)
#define forit(it,v) for(typeof((v).begin()) it = v.begin() ; it != (v).end() ; ++it)
#define eprintf(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define sz(a) ((int)(a).size())
#define all(a) (a).begin(),a.end()
#define two(i) (1LL<<(i))
typedef long long ll;
typedef vector<int> VI;
typedef pair<int, int> PII;
int n;
vector<ll> arr;
ll ans;
map<PII, ll> M;
void update(ll & ret, ll tmp) {
if (tmp == -1) return;
if (ret == -1) ret = tmp;
ret = min(ret, tmp);
}
/*
* memoization(mask, pos)
* Args:
* mask: if 2^i in mask it means arr[i] has been added a high bit before, and all lower bit(<=pos) can be considerd zero.
* pos: current check bit position, start from high to low
* Return:
* return -1 if not valid ans exists else return minimum addition sum
*/
int memoization(int mask, int pos) {
if (pos < 0) {
return 0;
}
PII state = mp(mask, pos);
if (M.find(state) != M.end()) {
return M[state];
}
ll &ret = M[state];
ret = -1;
int one_cnt = 0;
for (int i = 0; i < n; i++) {
if ( !(mask & two(i)) &&
(two(pos) & arr[i])) {
one_cnt ++;
}
}
if (one_cnt % 2 == 0) { // even, xor on this pos equals zero
ret = memoization(mask, pos - 1);
} else {
if (one_cnt == n) { //full odd bad state, do nothing
//pass
} else { //not full odd, choose one empty bit to place 1
for (int i = 0; i < n; i++) {
if ((mask & two(i)) //if number i has been added before, then it contain zero at pos
|| !(two(pos) & arr[i]) // or if number i has zero at pos and hasn't been added before
) {
ll candi = memoization(mask | two(i), pos - 1);
ll added = mask & two(i) ? two(pos) // number i has been added before, so we need extra two(pos) sum
//number i hasn't been added before, we need calc the new sum
//here we only consider bits in [0 .. pos]
: two(pos) - arr[i] % two(pos + 1);
if (candi >= 0) // legal result
update(ret, candi + added);
}
}
}
}
return ret;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("g.in", "r", stdin);
#endif
while (cin >> n) {
arr.clear();
for (int i = 0; i < n; i++) {
ll val;
cin >> val;
arr.push_back(val);
}
ll max_val = arr[0];
for (int i = 1; i < n; i++) max_val = max(max_val, arr[i]);
int max_pos = 0;
while (max_val) max_pos ++, max_val >>= 1;
max_pos ++;
//no adjust
M.clear();
ans = memoization(0, 31);
bool even_bit = true;
for (int i = max_pos; i >= 0; i--) {
int one_cnt = 0;
for (int j = 0; j < n; j++) one_cnt += (two(i) & arr[j]) > 0;
even_bit &= one_cnt % 2 == 0;
if (even_bit) {
for (int j = 0; j < n; j++) {
//arr[j] at pos i is empty, try add to 1
if (!(two(i) & arr[j])) {
ll backup = arr[j];
arr[j] = two(i);
//since previous pos all contain even one bits, we just start from current pos i
M.clear();
ll candi = memoization(0, i);
ll added = two(i) - backup % two(i);
if (candi >= 0)
update(ans, candi + added);
arr[j] = backup;
}
}
}
}
cout << ans << endl;
}
return 0;
}